Integrand size = 23, antiderivative size = 126 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {b^2 (6 a+5 b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 a^{7/2} (a+b)^{3/2} f}+\frac {(a-2 b) \sin (e+f x)}{a^3 f}-\frac {\sin ^3(e+f x)}{3 a^2 f}-\frac {b^3 \sin (e+f x)}{2 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )} \]
1/2*b^2*(6*a+5*b)*arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))/a^(7/2)/(a+b)^(3 /2)/f+(a-2*b)*sin(f*x+e)/a^3/f-1/3*sin(f*x+e)^3/a^2/f-1/2*b^3*sin(f*x+e)/a ^3/(a+b)/f/(a+b-a*sin(f*x+e)^2)
Time = 0.83 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {-\frac {3 b^2 (6 a+5 b) \left (\log \left (\sqrt {a+b}-\sqrt {a} \sin (e+f x)\right )-\log \left (\sqrt {a+b}+\sqrt {a} \sin (e+f x)\right )\right )}{(a+b)^{3/2}}+3 \sqrt {a} \left (3 a-8 b-\frac {4 b^3}{(a+b) (a+2 b+a \cos (2 (e+f x)))}\right ) \sin (e+f x)+a^{3/2} \sin (3 (e+f x))}{12 a^{7/2} f} \]
((-3*b^2*(6*a + 5*b)*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x]] - Log[Sqrt[a + b] + Sqrt[a]*Sin[e + f*x]]))/(a + b)^(3/2) + 3*Sqrt[a]*(3*a - 8*b - (4* b^3)/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))*Sin[e + f*x] + a^(3/2)*Sin[ 3*(e + f*x)])/(12*a^(7/2)*f)
Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4635, 300, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4635 |
\(\displaystyle \frac {\int \frac {\left (1-\sin ^2(e+f x)\right )^3}{\left (-a \sin ^2(e+f x)+a+b\right )^2}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (-\frac {\sin ^2(e+f x)}{a^2}+\frac {a-2 b}{a^3}+\frac {b^2 (3 a+2 b)-3 a b^2 \sin ^2(e+f x)}{a^3 \left (-a \sin ^2(e+f x)+a+b\right )^2}\right )d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2 (6 a+5 b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 a^{7/2} (a+b)^{3/2}}-\frac {b^3 \sin (e+f x)}{2 a^3 (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {(a-2 b) \sin (e+f x)}{a^3}-\frac {\sin ^3(e+f x)}{3 a^2}}{f}\) |
((b^2*(6*a + 5*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(7/2)* (a + b)^(3/2)) + ((a - 2*b)*Sin[e + f*x])/a^3 - Sin[e + f*x]^3/(3*a^2) - ( b^3*Sin[e + f*x])/(2*a^3*(a + b)*(a + b - a*Sin[e + f*x]^2)))/f
3.2.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 3.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {a \sin \left (f x +e \right )^{3}}{3}-\sin \left (f x +e \right ) a +2 \sin \left (f x +e \right ) b}{a^{3}}-\frac {b^{2} \left (-\frac {b \sin \left (f x +e \right )}{2 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}-\frac {\left (6 a +5 b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}}{f}\) | \(120\) |
default | \(\frac {-\frac {\frac {a \sin \left (f x +e \right )^{3}}{3}-\sin \left (f x +e \right ) a +2 \sin \left (f x +e \right ) b}{a^{3}}-\frac {b^{2} \left (-\frac {b \sin \left (f x +e \right )}{2 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}-\frac {\left (6 a +5 b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}}{f}\) | \(120\) |
risch | \(-\frac {i {\mathrm e}^{3 i \left (f x +e \right )}}{24 a^{2} f}-\frac {3 i {\mathrm e}^{i \left (f x +e \right )}}{8 a^{2} f}+\frac {i {\mathrm e}^{i \left (f x +e \right )} b}{a^{3} f}+\frac {3 i {\mathrm e}^{-i \left (f x +e \right )}}{8 a^{2} f}-\frac {i {\mathrm e}^{-i \left (f x +e \right )} b}{a^{3} f}+\frac {i {\mathrm e}^{-3 i \left (f x +e \right )}}{24 a^{2} f}+\frac {i b^{3} \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{a^{3} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, \left (a +b \right ) f \,a^{2}}+\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right ) f \,a^{3}}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, \left (a +b \right ) f \,a^{2}}-\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right ) f \,a^{3}}\) | \(433\) |
1/f*(-1/a^3*(1/3*a*sin(f*x+e)^3-sin(f*x+e)*a+2*sin(f*x+e)*b)-b^2/a^3*(-1/2 /(a+b)*b*sin(f*x+e)/(a*sin(f*x+e)^2-a-b)-1/2*(6*a+5*b)/(a+b)/(a*(a+b))^(1/ 2)*arctanh(a*sin(f*x+e)/(a*(a+b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (115) = 230\).
Time = 0.31 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.89 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {3 \, {\left (6 \, a b^{3} + 5 \, b^{4} + {\left (6 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4} + 2 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{5} - a^{4} b - 8 \, a^{3} b^{2} - 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{12 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}, -\frac {3 \, {\left (6 \, a b^{3} + 5 \, b^{4} + {\left (6 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) - {\left (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4} + 2 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{5} - a^{4} b - 8 \, a^{3} b^{2} - 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}\right ] \]
[1/12*(3*(6*a*b^3 + 5*b^4 + (6*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*sqrt(a^2 + a*b)*sin(f*x + e) - 2*a - b) /(a*cos(f*x + e)^2 + b)) + 2*(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^4 + 2*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^4 + 2*(2*a^5 - a^4*b - 8*a^3*b^ 2 - 5*a^2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f* cos(f*x + e)^2 + (a^6*b + 2*a^5*b^2 + a^4*b^3)*f), -1/6*(3*(6*a*b^3 + 5*b^ 4 + (6*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^ 2 - a*b)*sin(f*x + e)/(a + b)) - (4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a* b^4 + 2*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^4 + 2*(2*a^5 - a^4*b - 8*a^ 3*b^2 - 5*a^2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2 )*f*cos(f*x + e)^2 + (a^6*b + 2*a^5*b^2 + a^4*b^3)*f)]
Timed out. \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {6 \, b^{3} \sin \left (f x + e\right )}{a^{5} + 2 \, a^{4} b + a^{3} b^{2} - {\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} + \frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} a}} + \frac {4 \, {\left (a \sin \left (f x + e\right )^{3} - 3 \, {\left (a - 2 \, b\right )} \sin \left (f x + e\right )\right )}}{a^{3}}}{12 \, f} \]
-1/12*(6*b^3*sin(f*x + e)/(a^5 + 2*a^4*b + a^3*b^2 - (a^5 + a^4*b)*sin(f*x + e)^2) + 3*(6*a*b^2 + 5*b^3)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*s in(f*x + e) + sqrt((a + b)*a)))/((a^4 + a^3*b)*sqrt((a + b)*a)) + 4*(a*sin (f*x + e)^3 - 3*(a - 2*b)*sin(f*x + e))/a^3)/f
Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \, b^{3} \sin \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}} - \frac {3 \, {\left (6 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {-a^{2} - a b}} - \frac {2 \, {\left (a^{4} \sin \left (f x + e\right )^{3} - 3 \, a^{4} \sin \left (f x + e\right ) + 6 \, a^{3} b \sin \left (f x + e\right )\right )}}{a^{6}}}{6 \, f} \]
1/6*(3*b^3*sin(f*x + e)/((a^4 + a^3*b)*(a*sin(f*x + e)^2 - a - b)) - 3*(6* a*b^2 + 5*b^3)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^4 + a^3*b)*sqrt (-a^2 - a*b)) - 2*(a^4*sin(f*x + e)^3 - 3*a^4*sin(f*x + e) + 6*a^3*b*sin(f *x + e))/a^6)/f
Time = 19.79 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (6\,a+5\,b\right )}{2\,a^{7/2}\,f\,{\left (a+b\right )}^{3/2}}-\frac {{\sin \left (e+f\,x\right )}^3}{3\,a^2\,f}-\frac {b^3\,\sin \left (e+f\,x\right )}{2\,f\,\left (a+b\right )\,\left (-a^4\,{\sin \left (e+f\,x\right )}^2+a^4+b\,a^3\right )}-\frac {\sin \left (e+f\,x\right )\,\left (\frac {2\,\left (a+b\right )}{a^3}-\frac {3}{a^2}\right )}{f} \]